Answered

Note: Enter your answer and show all the steps that you use to solve this problem in the space providedA shot put is thrown upward with a velocity of 35 ft/sec. at a height of 4 ft. and an angle of 40°. How longwill it take for the shot put to be a horizontal distance of 40 ft. from the person throwing it? Accelerationdue to gravity is 32 ft./s². Round your answer to the nearest hundredth.

Note Enter your answer and show all the steps that you use to solve this problem in the space providedA shot put is thrown upward with a velocity of 35 ftsec at class=


Answer :

Given

Initial velocity = 35 ft/sec

Initial height of shot put from horizontal , h' = 4 ft

Angle = 40 degree

Find

Time taken for the shot put to be a horizontal distance of 40 ft. from the person throwing it

Explanation

First we find the horizontal component of velocity

Horizontal component of velocity is

[tex]\begin{gathered} u_x=u\cos\theta \\ u_x=35\cos40\degree \\ u_x=26.812\frac{ft}{sec} \end{gathered}[/tex]

the time taken by the shot put to travel a horizontal distance of 40 ft. from the person throwing it is

[tex]\begin{gathered} t=\frac{40}{u_x} \\ \\ t=\frac{40}{26.812}=1.4918\approx1.50sec \end{gathered}[/tex]

Final Answer

Therefore , the time taken by the shot put to travel a horizontal distance of 40 ft. from the person throwing it is 1.50 seconds

View image SuleymaM747086

Other Questions